Question

Suppose you start with a solution of red dye #40 that is 3.1 ✕ 10−5 M. If you do four successive volumetric dilutions pipetting 1.00 mL of solution and diluting with water in a 40.00 mL volumetric flask, what is the molarity of the final dilution?

Answer 1

Answer:

1.3 × 10⁻¹¹ M

Explanation:

We are going to do 4 successive dilutions. In each dilution, we will apply the dilution rule.

C₁.V₁=C₂.V₂

where,

C₁ and V₁ are concentration and volume of the initial state

C₂ and V₂ are concentration and volume of the final state

First dilution

C₁ = 3.1 × 10⁻⁵ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

$C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{3.1\times 10^{-5}M \times 1.00mL }{40.00mL} =7.8 \times 10^{-7}M$

Second dilution

C₁ = 7.8 × 10⁻⁷ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

$C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{7.8 \times 10^{-7}M \times 1.00mL }{40.00mL} =2.0 \times 10^{-8}M$

Third dilution

C₁ = 2.0 × 10⁻⁸ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

$C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{2.0 \times 10^{-8}M \times 1.00mL }{40.00mL} =5.0 \times 10^{-10}M$

Fourth dilution

C₁ = 5.0 × 10⁻¹⁰ M V₁ = 1.00 mL C₂ = ? V₂ = 40.00mL

$C_{2}=\frac{C_{1}.V_{1}}{V_{2}} =\frac{5.0 \times 10^{-10}M \times 1.00mL }{40.00mL} =1.3 \times 10^{-11}M$